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a(n) = ( 1/1 + 1/3 + 1/5 + ... + 1/(2*n-1) )*LCM(1, 3, 5, ..., 2*n-1).
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%I #67 Aug 14 2022 15:27:27

%S 1,4,23,176,563,6508,88069,91072,1593269,31037876,31730711,744355888,

%T 3788707301,11552032628,340028535787,10686452707072,10823198495797,

%U 10952130239452,409741429887649,414022624965424,17141894231615609,743947082888833412,750488463554668427

%N a(n) = ( 1/1 + 1/3 + 1/5 + ... + 1/(2*n-1) )*LCM(1, 3, 5, ..., 2*n-1).

%C Or, numerator of 1/1 + 1/3 + ... + 1/(2n-1) up to a(38).

%C Following similar remark by _T. D. Noe_ in A025547, this coincides with f(n) = numerator of 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n-1) iff n <= 38. But a(39) = 18048708369314455836683437302413, f(39) = 1640791669937677803334857936583. Note that f(n) = numerator(digamma(n+1/2)/2 + log(2) + euler_gamma/2). - _Paul Barry_, Aug 19 2005 [See A350669(n-1).]

%H Georg Fischer, <a href="/A025550/b025550.txt">Table of n, a(n) for n = 1..200</a> (first 39 terms from Jean-François Alcover)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/JeepProblem.html">Jeep Problem</a>

%p a:= n-> (f-> add(1/p, p=f)*ilcm(f[]))([2*i-1$i=1..n]):

%p seq(a(n), n=1..40); # _Alois P. Heinz_, Apr 16 2015

%t Table[(Total[1/Range[1,2n-1,2]])LCM@@Range[1,2n-1,2],{n,30}] (* _Harvey P. Dale_, Sep 09 2020 *)

%o (Haskell)

%o a025550 n = numerator $ sum $ map (1 %) $ take n [1, 3 ..]

%o -- _Reinhard Zumkeller_, Jan 22 2012

%o (PARI) a(n)=my(v=vector(n,i,2*i-1));sum(i=1,#v,1/v[i])*lcm(v) \\ _Charles R Greathouse IV_, Feb 28 2013

%o (Magma) [&+[1/d: d in i]*Lcm(i) where i is [1..2*n-1 by 2]: n in [1..21]]; // _Bruno Berselli_, Apr 16 2015

%Y Cf. A025547, A075135, A002428, A350669.

%K nonn,easy,nice,frac

%O 1,2

%A _Clark Kimberling_

%E Value of a(39) corrected by _Jean-François Alcover_, Apr 16 2015